Only using digits 0 to 9 (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) once, can you form three numbers *a, b *and *c *so that the sixth power of *c *is equal to the sum of *a *and *b*?

## Solution

in essence, the problem asked you to find the solution of *c*^{6} = *a* + *b* by using digits 0-9 once. Since only ten digits are used for this problem, *a *would most likely not contain more than sеven digits*. So, 4 < c *< 15. Now any sixth power is congruent to 0 or to 1 modulo 9, so *a *+ *b *≡ 0 or 1 mod 9. Sincе the sum of thе ten digits is 45, it follows that *c *is еither 8 or 9. Therefore, *b *has three digits. With some trial and error, I found three basic solutions.

261709 + 435 = 8^{6}

261790 + 354 = 8^{6}

530827 + 614 = 8^{6}

In each solution, the hundred’s digit of *a *and *b *can be interchanged without changing the value of *a* + *b. *The same goes for the ten’s digits and for the unit’s digits. Moreover, *a *and *b *are interchangeable. Hence, the problem has 2³ × 2 × 3 or 48 solutions.

Egad. If I could even understand this, it would be my sixth power….

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So you already have five powers 😀

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Are you kidding…..:)

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Still trying to wrap my head around how you even narrowed it down.

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Doing lots of puzzles everyday helped a lot :).

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Yes, absolutely.

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Way over my head.

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The third basic solution utilizes 8 twice in violation of the proposed problem.

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Also, all three listed solutions equal 8 to the sixth power, yet the first number of the third posited solution is far greater than 262,144.

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